Get all possible palindromes from a given string

Question

A substring is a contiguous range of characters within a string.

Now I need to find out how many substring that can be re-arranged that can form a palindrome.

For example: For input - aabb

a
aa
aab (because after re-arranging it becomes aba)
aabb (because after re-arranging it becomes abba)
a
abb (because after re-arranging it becomes bab)
b
bb
b

So we have 9 substring palindromes.

Here is the code I tried:

public static int getPalindromeCount(String str) {
    // all single characters are treated as palindrome
    int count = str.length();

    // Get all sub strings
    List<String> subs = new ArrayList<>();
    subString(str, subs);

    for (String sub : subs) {
        String rev = new StringBuilder(sub).reverse().toString();

        if (rev.equals(sub)) {
            System.out.println(sub);
            count++;
        } else {
            boolean valid = isPalindrome(sub);
            System.out.println(sub + " : " + valid);
            if (valid) {
                count++;
            }
        }
    }
    return count;
}

// Check if substring can form a Palindrome
private static boolean isPalindrome(String input) {
    Set<Character> oddChars = new HashSet<>();

    for (char c : input.toCharArray()) {
        if (!oddChars.add(c)) {
            oddChars.remove(c);
        }
    }
    return oddChars.size() <= 1;
}

// Get all substrings
private static void subString(String input, List<String> list) {
    for (int i = 0; i < input.length(); i++) {
        for (int j = i + 2; j <= input.length(); j++) {
            list.add(input.substring(i, j));
        }
    }
}

The method isPalindrome part of logic I have took from this post Check if a permutation of a string can become a palindrome

This code is working fine, but it is failing with time out errors.

I am not sure what are the inputs for which this is failing as they are hidden in my hackerrank challenge.

Edit:

I have modified my getPalidromeCount method to check for how many odd number of letters are there in input to decide palindrome count.

This is based on comment on this post:

Hint: A palindrome consists of all letters of even count or all letters of even count with one letter of odd count(the middle character). Now, you could count possible palindromes easily. – vivek_23

public static int getPalindromeCount(String str) {
    List<Integer> list = new ArrayList<>(strToEvaluate.size());
    for (String str : strToEvaluate) {
        int count = str.length();

        List<String> subs = new ArrayList<>();
        subString(str, subs);
        for (String sub : subs) {
            Map<Character, Integer> map = new HashMap<>();
            for (int i = 0; i < sub.length(); i++) {
                char c = sub.charAt(i);
                map.put(c, map.getOrDefault(c, 0) + 1);
            }
            int odds = 0;
            for (char key : map.keySet()) {
                if (map.get(key) % 2 != 0) {
                    odds++;
                    if (odds > 1) {
                        break;
                    }
                }
            }
            if (odds <= 1) {
                System.out.println(sub);
                count++;
            }

            list.add(count);
        }
    }
    return list;
}

But still I am seeing timeout errors. I am not using the isPalindrome method in this logic.

Answer 1

There are n(n+1)/2 possible substrings and for each substring you check whether it can be re-arranged so that it forms a palindrome in O(k) where k is length of given substring, let's think if it is necessary to parse each substring separately.

Hint:

Let's say you have substring from index p to k, what can you say about substring from index p to k + 1. Is it really necessary to parse this extended substring separately?

Answer 2

• My assumption is that string consists of lowercase letters. If not, you could increase the hash size in the below array to accommodate ASCII values or better use a map for UTF-8 characters.
• Instead of collecting all substrings and then individually checking each one(which would take O(n²) space, you could just iterate over them along the way, increase the current character count and check for palindromeness as shown below.

Code:

 class Solution{
  public static long getPalindromeCount(String s) {
    long cnt = 0,len = s.length();    
    for(int i=0;i<len;++i){
       int[] hash = new int[26];
       cnt++; // since 1 character is palindrome anyway
       for(int j=i+1;j<len;++j){
          hash[s.charAt(j)-'a']++;
          if(palindromePossible(hash)) cnt++; 
       }
    }
    return cnt;
  }

  private static boolean palindromePossible(int[] hash){
    int odd_cnt = 0;
    for(int i=0;i<hash.length;++i){
      if(hash[i] % 2 != 0) odd_cnt++;
    }
    return odd_cnt < 2;
  }

  public static void main(String[] args){
    System.out.println(getPalindromeCount("aabb"));
  }
}